only 2nd question plz help me related to coordinate geometry , distance formula
Answers
Answer:
25 sq, units
Step-by-step explanation:
Please see the attached figure. Roughly drawn based on the points given.
Construction: Join B & D to form a diagonal of the quadrilateral.
Area of Quadrilateral ABCD = Area of ΔABD + Area of ΔBDC
Now to find area of triangle ABD
Area of triangle = 1/2[x₁(y₂ -y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
x₁ = 1 , y₁ = 0
x₂ = 5 , y₂ = 3
x₃ = -2, y₃ = 4.
Substitute the values in above formula,
Area of ΔABD = 1/2[1( 3-4)+ 5(4-0)+(-2)(0-3)]
= 1/2[ -1 + 20 + 6] = 25/2 sq. units
Now to find area of triangle BDC
Area of triangle = 1/2[x₁(y₂ -y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]
x₁ = 5 , y₁ = 3
x₂ = -2 , y₂ = 4
x₃ = 2, y₃ = 7.
Substitute the values in above formula,
Area of ΔBDC = 1/2[5(4-7)+ (-2)(7-3)+2(3-4)]
= 1/2[-15 - 8 -2] = = -25/2.
Since Area of triangle cannot be negative,
The area of ΔBDC = 25/2 sq. units
Thus the area of quadrilateral ABCD = Area of ΔABD + Area of ΔBDC
= 25/2 + 25/2 = 50/2 = 25 sq. units
